When 0.005kg of ice at 0°C is added to 0.02kg of warm water at 30°C the final temperature attained is 8°C.Find the specific latent heat of fusion of ice

Answer 1

To find the specific latent heat of fusion of ice, we need to use the principle of conservation of energy. The heat lost by the warm water will be equal to the heat gained by the ice.

Let:

• \( m_i = 0.005 , \text{kg} \) (mass of ice)
• \( m_w = 0.02 , \text{kg} \) (mass of warm water)
• \( c_w = 4184 , \text{J/kg°C} \) (specific heat capacity of water)
• \( L_f \) = specific latent heat of fusion of ice (which we need to find)
• Initial temperature of ice, \( T_{i} = 0 , °C \)
• Initial temperature of water, \( T_{w} = 30 , °C \)
• Final temperature, \( T_f = 8 , °C \)

1. Heat gained by the ice: The heat gained by the ice includes the heat to melt the ice and the heat to warm the resulting water to the final temperature.

   When ice melts: \[ Q_{\text{ice melt}} = m_i L_f \]

   Then, the melted ice (which is now water at \( 0°C \)) warms up to the final temperature: \[ Q_{\text{warming}} = m_i c_w (T_f - T_i) = m_i c_w (8 - 0) = m_i c_w \times 8 \]

   Therefore, the total heat gained by the ice is: \[ Q_{\text{gain}} = m_i L_f + m_i c_w \times 8 \]

2. Heat lost by the warm water: The water cools from \( 30°C \) to \( 8°C \): \[ Q_{\text{loss}} = m_w c_w (T_w - T_f) = m_w c_w (30 - 8) = m_w c_w \times 22 \]

3. Setting up the equation: According to the conservation of energy: \[ Q_{\text{gain}} = Q_{\text{loss}} \] Plugging in the expressions we have: \[ m_i L_f + m_i c_w \times 8 = m_w c_w \times 22 \]

4. Substituting the known values: \[ 0.005 L_f + 0.005 \times 4184 \times 8 = 0.02 \times 4184 \times 22 \]

   Calculate \( Q_{\text{warming}} \) for the ice: \[ 0.005 \times 4184 \times 8 = 167.36 , \text{J} \]

   Calculate \( Q_{\text{loss}} \) for the water: \[ 0.02 \times 4184 \times 22 = 1841.44 , \text{J} \]

   Now we have the equation: \[ 0.005 L_f + 167.36 = 1841.44 \]

   Rearranging for \( L_f \): \[ 0.005 L_f = 1841.44 - 167.36 \] \[ 0.005 L_f = 1674.08 \]

5. Solving for \( L_f \): \[ L_f = \frac{1674.08}{0.005} = 334816 , \text{J/kg} \]

Thus, the specific latent heat of fusion of ice is approximately \( 334816 , \text{J/kg} \) or about \( 334.8 , \text{kJ/kg} \).